Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 54

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\[\textbf{а)}\ f(x) = e^{x^{3}};\ \ \ x \in R\]

\[y = e^{u};\ \ u = x^{3};\ \]

\[f^{'}(x) = \left( e^{u} \right)_{u}^{'} \cdot \left( x^{3} \right)_{x}^{'} =\]

\[= e^{u} \cdot 3x^{2} = 3x^{2}e^{x^{3}}.\]

\[\textbf{б)}\ \ f(x) = e^{- x^{4}};\ \ x \in R\]

\[y = e^{u};\ \ u = - x^{4};\ \]

\[f^{'}(x) = \left( e^{u} \right)_{u}^{'} \cdot \left( {- x}^{4} \right)_{x}^{'} =\]

\[= e^{u} \cdot \left( - 4x^{3} \right) = - 4x^{3}e^{{- x}^{4}}.\]

\[\textbf{в)}\ f(x) = 3^{x^{3}};\ \ x \in R\]

\[y = 3^{u};\ \ u = x^{3};\ \]

\[f^{'}(x) = \left( 3^{u} \right)_{u}^{'} \cdot \left( x^{3} \right)_{x}^{'} =\]

\[= 3^{u}\ln 3 \cdot 3x^{2} = 3x^{2}3^{x^{3}}\ln 3.\]

\[\textbf{г)}\ f(x) = 5^{- x^{4}};\ \ x \in R\]

\[y = 5^{u};\ \ u = - x^{4};\ \]

\[f^{'}(x) = \left( 5^{u} \right)_{u}^{'} \cdot \left( {- x}^{4} \right)_{x}^{'} =\]

\[\text{=}5^{u}\ln 5 \cdot \left( - 4x^{3} \right) =\]

\[= - 4x^{3}5^{{- x}^{4}}\ln 5.\]

\[\textbf{д)}\ y = e^{\sin x};\ \ x \in R\]

\[y = e^{u};\ \ u = \sin x;\ \]

\[f^{'}(x) = \left( e^{u} \right)_{u}^{'} \cdot \left( \sin x \right)_{x}^{'} =\]

\[= e^{u} \cdot \cos x = e^{\sin x} \cdot \cos x.\]

\[\textbf{е)}\ f(x) = 9^{\cos x};\ \ x \in R\]

\[y = 9^{u};\ \ u = \cos x;\ \]

\[f^{'}(x) = \left( 9^{u} \right)_{u}^{'} \cdot \left( \cos x \right)_{x}^{'} =\]

\[= 9^{u}\ln 9 \cdot {( - sin}x) =\]

\[= - \sin x \cdot 9^{\cos x}\ln 9.\]